Finite presentation of a cyclic group












1












$begingroup$


I'm struggling to understand group presentation.



There is a theorem that says, every group $G$ is the image of a suitable free group F (free upon a set $X$), and there must exist a homomorphism $pi$ from F to G, so by first isomorphism theorem, $G$ is isomorphic to the quotient group $F/ker(pi)$.



Question 1, I know we can denote $G$ as $<X mid R>$ where $X$ is the generating set of $F$. Is $R$ the generating set of $ker(pi)$? What about the normal closure of $R$? I am really confused here. Basically I want a concrete example demonstrating the difference between $R$, normal closure of $R$, and $ker(pi)$.



Question 2, $Z_7={0,1,2,3,4,5,6}$ is a cyclic group generated by ${1}$, which can also be written as $<x mid x^7>$. It seems to be the case that the generating set of the free group $F$ is $X={x}$, and $R={x^7}$. But then $G=F/ker(pi)$, what exactly is $F$ and $pi$ in this example?






abstract-algebra group-theory group-presentation







share|cite|improve this question











$endgroup$








  • 4




    $begingroup$
    I'm the langle rangle fairy, here to let you know that $langle, rangle$ plays nicer with TeX than <, > does :)
    $endgroup$
    – Patrick Stevens
    Apr 18 '17 at 6:37










  • $begingroup$
    Well this is the notation used in my class.
    $endgroup$
    – W.Scott
    Apr 18 '17 at 7:02










  • $begingroup$
    Really? They use $<X mid R>$ rather than $langle X mid R rangle$? That's unusual. $langle, rangle$ are angle brackets; $<, >$ are less than/greater than signs.
    $endgroup$
    – Patrick Stevens
    Apr 18 '17 at 7:09












  • $begingroup$
    Sorry my bad. Yes you are right. Can you please give me some insight to my two questions? Especially the first one.
    $endgroup$
    – W.Scott
    Apr 18 '17 at 7:12
















1












$begingroup$


I'm struggling to understand group presentation.



There is a theorem that says, every group $G$ is the image of a suitable free group F (free upon a set $X$), and there must exist a homomorphism $pi$ from F to G, so by first isomorphism theorem, $G$ is isomorphic to the quotient group $F/ker(pi)$.



Question 1, I know we can denote $G$ as $<X mid R>$ where $X$ is the generating set of $F$. Is $R$ the generating set of $ker(pi)$? What about the normal closure of $R$? I am really confused here. Basically I want a concrete example demonstrating the difference between $R$, normal closure of $R$, and $ker(pi)$.



Question 2, $Z_7={0,1,2,3,4,5,6}$ is a cyclic group generated by ${1}$, which can also be written as $<x mid x^7>$. It seems to be the case that the generating set of the free group $F$ is $X={x}$, and $R={x^7}$. But then $G=F/ker(pi)$, what exactly is $F$ and $pi$ in this example?






abstract-algebra group-theory group-presentation







share|cite|improve this question











$endgroup$








  • 4




    $begingroup$
    I'm the langle rangle fairy, here to let you know that $langle, rangle$ plays nicer with TeX than <, > does :)
    $endgroup$
    – Patrick Stevens
    Apr 18 '17 at 6:37










  • $begingroup$
    Well this is the notation used in my class.
    $endgroup$
    – W.Scott
    Apr 18 '17 at 7:02










  • $begingroup$
    Really? They use $<X mid R>$ rather than $langle X mid R rangle$? That's unusual. $langle, rangle$ are angle brackets; $<, >$ are less than/greater than signs.
    $endgroup$
    – Patrick Stevens
    Apr 18 '17 at 7:09












  • $begingroup$
    Sorry my bad. Yes you are right. Can you please give me some insight to my two questions? Especially the first one.
    $endgroup$
    – W.Scott
    Apr 18 '17 at 7:12














1












1








1





$begingroup$


I'm struggling to understand group presentation.



There is a theorem that says, every group $G$ is the image of a suitable free group F (free upon a set $X$), and there must exist a homomorphism $pi$ from F to G, so by first isomorphism theorem, $G$ is isomorphic to the quotient group $F/ker(pi)$.



Question 1, I know we can denote $G$ as $<X mid R>$ where $X$ is the generating set of $F$. Is $R$ the generating set of $ker(pi)$? What about the normal closure of $R$? I am really confused here. Basically I want a concrete example demonstrating the difference between $R$, normal closure of $R$, and $ker(pi)$.



Question 2, $Z_7={0,1,2,3,4,5,6}$ is a cyclic group generated by ${1}$, which can also be written as $<x mid x^7>$. It seems to be the case that the generating set of the free group $F$ is $X={x}$, and $R={x^7}$. But then $G=F/ker(pi)$, what exactly is $F$ and $pi$ in this example?






abstract-algebra group-theory group-presentation







share|cite|improve this question











$endgroup$




I'm struggling to understand group presentation.



There is a theorem that says, every group $G$ is the image of a suitable free group F (free upon a set $X$), and there must exist a homomorphism $pi$ from F to G, so by first isomorphism theorem, $G$ is isomorphic to the quotient group $F/ker(pi)$.



Question 1, I know we can denote $G$ as $<X mid R>$ where $X$ is the generating set of $F$. Is $R$ the generating set of $ker(pi)$? What about the normal closure of $R$? I am really confused here. Basically I want a concrete example demonstrating the difference between $R$, normal closure of $R$, and $ker(pi)$.



Question 2, $Z_7={0,1,2,3,4,5,6}$ is a cyclic group generated by ${1}$, which can also be written as $<x mid x^7>$. It seems to be the case that the generating set of the free group $F$ is $X={x}$, and $R={x^7}$. But then $G=F/ker(pi)$, what exactly is $F$ and $pi$ in this example?






abstract-algebra group-theory group-presentation




abstract-algebra group-theory group-presentation






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 10 '18 at 23:39









Shaun

9,268113684




9,268113684










asked Apr 18 '17 at 6:28









W.ScottW.Scott

1037




1037








  • 4




    $begingroup$
    I'm the langle rangle fairy, here to let you know that $langle, rangle$ plays nicer with TeX than <, > does :)
    $endgroup$
    – Patrick Stevens
    Apr 18 '17 at 6:37










  • $begingroup$
    Well this is the notation used in my class.
    $endgroup$
    – W.Scott
    Apr 18 '17 at 7:02










  • $begingroup$
    Really? They use $<X mid R>$ rather than $langle X mid R rangle$? That's unusual. $langle, rangle$ are angle brackets; $<, >$ are less than/greater than signs.
    $endgroup$
    – Patrick Stevens
    Apr 18 '17 at 7:09












  • $begingroup$
    Sorry my bad. Yes you are right. Can you please give me some insight to my two questions? Especially the first one.
    $endgroup$
    – W.Scott
    Apr 18 '17 at 7:12














  • 4




    $begingroup$
    I'm the langle rangle fairy, here to let you know that $langle, rangle$ plays nicer with TeX than <, > does :)
    $endgroup$
    – Patrick Stevens
    Apr 18 '17 at 6:37










  • $begingroup$
    Well this is the notation used in my class.
    $endgroup$
    – W.Scott
    Apr 18 '17 at 7:02










  • $begingroup$
    Really? They use $<X mid R>$ rather than $langle X mid R rangle$? That's unusual. $langle, rangle$ are angle brackets; $<, >$ are less than/greater than signs.
    $endgroup$
    – Patrick Stevens
    Apr 18 '17 at 7:09












  • $begingroup$
    Sorry my bad. Yes you are right. Can you please give me some insight to my two questions? Especially the first one.
    $endgroup$
    – W.Scott
    Apr 18 '17 at 7:12








4




4




$begingroup$
I'm the langle rangle fairy, here to let you know that $langle, rangle$ plays nicer with TeX than <, > does :)
$endgroup$
– Patrick Stevens
Apr 18 '17 at 6:37




$begingroup$
I'm the langle rangle fairy, here to let you know that $langle, rangle$ plays nicer with TeX than <, > does :)
$endgroup$
– Patrick Stevens
Apr 18 '17 at 6:37












$begingroup$
Well this is the notation used in my class.
$endgroup$
– W.Scott
Apr 18 '17 at 7:02




$begingroup$
Well this is the notation used in my class.
$endgroup$
– W.Scott
Apr 18 '17 at 7:02












$begingroup$
Really? They use $<X mid R>$ rather than $langle X mid R rangle$? That's unusual. $langle, rangle$ are angle brackets; $<, >$ are less than/greater than signs.
$endgroup$
– Patrick Stevens
Apr 18 '17 at 7:09






$begingroup$
Really? They use $<X mid R>$ rather than $langle X mid R rangle$? That's unusual. $langle, rangle$ are angle brackets; $<, >$ are less than/greater than signs.
$endgroup$
– Patrick Stevens
Apr 18 '17 at 7:09














$begingroup$
Sorry my bad. Yes you are right. Can you please give me some insight to my two questions? Especially the first one.
$endgroup$
– W.Scott
Apr 18 '17 at 7:12




$begingroup$
Sorry my bad. Yes you are right. Can you please give me some insight to my two questions? Especially the first one.
$endgroup$
– W.Scott
Apr 18 '17 at 7:12










1 Answer
1






active

oldest

votes


















2












$begingroup$

In the cyclic group $C_7$ with presentation $langle x mid x^7 rangle$, $F$ is the free group of rank $1$ with a single generator $x$ (so $F$ is an infinite cyclic group), $R = { x^7 }$, and $ker pi = langle R rangle =langle R^F rangle$ (the normal closure of $R$ in $F$). So in this example, $langle R rangle$ is equal to its normal closure. But that is not usually true.



For another example, consider $G = langle x,y mid r rangle$ with $r = [x,y] = xyx^{-1}y^{-1}$, which is a presentation of the free abelian group of rank $2$. Then $G = F/langle R^F rangle$ with $F$ a free group of rank $2$ and generators $x,y$, and $R = { r }$.



So $langle R rangle$ is a cyclic subgroup of $F$, but it is not normal. It's normal closure $langle R^F rangle$ is not even finitely generated. It contains all of the conjuagtes of $r$,like $[x^i,y^j]$ for all $i,j in {mathbb Z}$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    How exactly do you compute normal closure of $R$?
    $endgroup$
    – W.Scott
    Apr 18 '17 at 8:14










  • $begingroup$
    $langle R^F rangle$ is the by definition the subgroup $langle frf^{-1} : r in R, f in F rangle$ of $F$. It is not usually finitely generated. What exactly do you mean by "compute"?
    $endgroup$
    – Derek Holt
    Apr 18 '17 at 9:48










  • $begingroup$
    So is the normal closure of $R$ always equal to the kernel of $pi$? Because on one hand, $G$ is isomorphic to $F/ker(pi)$, while on the other hand $G$ is isomorphic to $F/N$ where $N$ is the normal closure of $R$.
    $endgroup$
    – W.Scott
    Apr 18 '17 at 11:45












  • $begingroup$
    And for the $Z_7$ example, how can you show that $<R>=<R^F>$?
    $endgroup$
    – W.Scott
    Apr 18 '17 at 11:49












  • $begingroup$
    Yes, with your notation $N = {rm ker} pi$. In the $Z_7$ example $F$ is cyclic and hence abelian.
    $endgroup$
    – Derek Holt
    Apr 18 '17 at 12:13











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1 Answer
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oldest

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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes









2












$begingroup$

In the cyclic group $C_7$ with presentation $langle x mid x^7 rangle$, $F$ is the free group of rank $1$ with a single generator $x$ (so $F$ is an infinite cyclic group), $R = { x^7 }$, and $ker pi = langle R rangle =langle R^F rangle$ (the normal closure of $R$ in $F$). So in this example, $langle R rangle$ is equal to its normal closure. But that is not usually true.



For another example, consider $G = langle x,y mid r rangle$ with $r = [x,y] = xyx^{-1}y^{-1}$, which is a presentation of the free abelian group of rank $2$. Then $G = F/langle R^F rangle$ with $F$ a free group of rank $2$ and generators $x,y$, and $R = { r }$.



So $langle R rangle$ is a cyclic subgroup of $F$, but it is not normal. It's normal closure $langle R^F rangle$ is not even finitely generated. It contains all of the conjuagtes of $r$,like $[x^i,y^j]$ for all $i,j in {mathbb Z}$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    How exactly do you compute normal closure of $R$?
    $endgroup$
    – W.Scott
    Apr 18 '17 at 8:14










  • $begingroup$
    $langle R^F rangle$ is the by definition the subgroup $langle frf^{-1} : r in R, f in F rangle$ of $F$. It is not usually finitely generated. What exactly do you mean by "compute"?
    $endgroup$
    – Derek Holt
    Apr 18 '17 at 9:48










  • $begingroup$
    So is the normal closure of $R$ always equal to the kernel of $pi$? Because on one hand, $G$ is isomorphic to $F/ker(pi)$, while on the other hand $G$ is isomorphic to $F/N$ where $N$ is the normal closure of $R$.
    $endgroup$
    – W.Scott
    Apr 18 '17 at 11:45












  • $begingroup$
    And for the $Z_7$ example, how can you show that $<R>=<R^F>$?
    $endgroup$
    – W.Scott
    Apr 18 '17 at 11:49












  • $begingroup$
    Yes, with your notation $N = {rm ker} pi$. In the $Z_7$ example $F$ is cyclic and hence abelian.
    $endgroup$
    – Derek Holt
    Apr 18 '17 at 12:13
















2












$begingroup$

In the cyclic group $C_7$ with presentation $langle x mid x^7 rangle$, $F$ is the free group of rank $1$ with a single generator $x$ (so $F$ is an infinite cyclic group), $R = { x^7 }$, and $ker pi = langle R rangle =langle R^F rangle$ (the normal closure of $R$ in $F$). So in this example, $langle R rangle$ is equal to its normal closure. But that is not usually true.



For another example, consider $G = langle x,y mid r rangle$ with $r = [x,y] = xyx^{-1}y^{-1}$, which is a presentation of the free abelian group of rank $2$. Then $G = F/langle R^F rangle$ with $F$ a free group of rank $2$ and generators $x,y$, and $R = { r }$.



So $langle R rangle$ is a cyclic subgroup of $F$, but it is not normal. It's normal closure $langle R^F rangle$ is not even finitely generated. It contains all of the conjuagtes of $r$,like $[x^i,y^j]$ for all $i,j in {mathbb Z}$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    How exactly do you compute normal closure of $R$?
    $endgroup$
    – W.Scott
    Apr 18 '17 at 8:14










  • $begingroup$
    $langle R^F rangle$ is the by definition the subgroup $langle frf^{-1} : r in R, f in F rangle$ of $F$. It is not usually finitely generated. What exactly do you mean by "compute"?
    $endgroup$
    – Derek Holt
    Apr 18 '17 at 9:48










  • $begingroup$
    So is the normal closure of $R$ always equal to the kernel of $pi$? Because on one hand, $G$ is isomorphic to $F/ker(pi)$, while on the other hand $G$ is isomorphic to $F/N$ where $N$ is the normal closure of $R$.
    $endgroup$
    – W.Scott
    Apr 18 '17 at 11:45












  • $begingroup$
    And for the $Z_7$ example, how can you show that $<R>=<R^F>$?
    $endgroup$
    – W.Scott
    Apr 18 '17 at 11:49












  • $begingroup$
    Yes, with your notation $N = {rm ker} pi$. In the $Z_7$ example $F$ is cyclic and hence abelian.
    $endgroup$
    – Derek Holt
    Apr 18 '17 at 12:13














2












2








2





$begingroup$

In the cyclic group $C_7$ with presentation $langle x mid x^7 rangle$, $F$ is the free group of rank $1$ with a single generator $x$ (so $F$ is an infinite cyclic group), $R = { x^7 }$, and $ker pi = langle R rangle =langle R^F rangle$ (the normal closure of $R$ in $F$). So in this example, $langle R rangle$ is equal to its normal closure. But that is not usually true.



For another example, consider $G = langle x,y mid r rangle$ with $r = [x,y] = xyx^{-1}y^{-1}$, which is a presentation of the free abelian group of rank $2$. Then $G = F/langle R^F rangle$ with $F$ a free group of rank $2$ and generators $x,y$, and $R = { r }$.



So $langle R rangle$ is a cyclic subgroup of $F$, but it is not normal. It's normal closure $langle R^F rangle$ is not even finitely generated. It contains all of the conjuagtes of $r$,like $[x^i,y^j]$ for all $i,j in {mathbb Z}$.






share|cite|improve this answer









$endgroup$



In the cyclic group $C_7$ with presentation $langle x mid x^7 rangle$, $F$ is the free group of rank $1$ with a single generator $x$ (so $F$ is an infinite cyclic group), $R = { x^7 }$, and $ker pi = langle R rangle =langle R^F rangle$ (the normal closure of $R$ in $F$). So in this example, $langle R rangle$ is equal to its normal closure. But that is not usually true.



For another example, consider $G = langle x,y mid r rangle$ with $r = [x,y] = xyx^{-1}y^{-1}$, which is a presentation of the free abelian group of rank $2$. Then $G = F/langle R^F rangle$ with $F$ a free group of rank $2$ and generators $x,y$, and $R = { r }$.



So $langle R rangle$ is a cyclic subgroup of $F$, but it is not normal. It's normal closure $langle R^F rangle$ is not even finitely generated. It contains all of the conjuagtes of $r$,like $[x^i,y^j]$ for all $i,j in {mathbb Z}$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Apr 18 '17 at 8:04









Derek HoltDerek Holt

53.7k53571




53.7k53571












  • $begingroup$
    How exactly do you compute normal closure of $R$?
    $endgroup$
    – W.Scott
    Apr 18 '17 at 8:14










  • $begingroup$
    $langle R^F rangle$ is the by definition the subgroup $langle frf^{-1} : r in R, f in F rangle$ of $F$. It is not usually finitely generated. What exactly do you mean by "compute"?
    $endgroup$
    – Derek Holt
    Apr 18 '17 at 9:48










  • $begingroup$
    So is the normal closure of $R$ always equal to the kernel of $pi$? Because on one hand, $G$ is isomorphic to $F/ker(pi)$, while on the other hand $G$ is isomorphic to $F/N$ where $N$ is the normal closure of $R$.
    $endgroup$
    – W.Scott
    Apr 18 '17 at 11:45












  • $begingroup$
    And for the $Z_7$ example, how can you show that $<R>=<R^F>$?
    $endgroup$
    – W.Scott
    Apr 18 '17 at 11:49












  • $begingroup$
    Yes, with your notation $N = {rm ker} pi$. In the $Z_7$ example $F$ is cyclic and hence abelian.
    $endgroup$
    – Derek Holt
    Apr 18 '17 at 12:13


















  • $begingroup$
    How exactly do you compute normal closure of $R$?
    $endgroup$
    – W.Scott
    Apr 18 '17 at 8:14










  • $begingroup$
    $langle R^F rangle$ is the by definition the subgroup $langle frf^{-1} : r in R, f in F rangle$ of $F$. It is not usually finitely generated. What exactly do you mean by "compute"?
    $endgroup$
    – Derek Holt
    Apr 18 '17 at 9:48










  • $begingroup$
    So is the normal closure of $R$ always equal to the kernel of $pi$? Because on one hand, $G$ is isomorphic to $F/ker(pi)$, while on the other hand $G$ is isomorphic to $F/N$ where $N$ is the normal closure of $R$.
    $endgroup$
    – W.Scott
    Apr 18 '17 at 11:45












  • $begingroup$
    And for the $Z_7$ example, how can you show that $<R>=<R^F>$?
    $endgroup$
    – W.Scott
    Apr 18 '17 at 11:49












  • $begingroup$
    Yes, with your notation $N = {rm ker} pi$. In the $Z_7$ example $F$ is cyclic and hence abelian.
    $endgroup$
    – Derek Holt
    Apr 18 '17 at 12:13
















$begingroup$
How exactly do you compute normal closure of $R$?
$endgroup$
– W.Scott
Apr 18 '17 at 8:14




$begingroup$
How exactly do you compute normal closure of $R$?
$endgroup$
– W.Scott
Apr 18 '17 at 8:14












$begingroup$
$langle R^F rangle$ is the by definition the subgroup $langle frf^{-1} : r in R, f in F rangle$ of $F$. It is not usually finitely generated. What exactly do you mean by "compute"?
$endgroup$
– Derek Holt
Apr 18 '17 at 9:48




$begingroup$
$langle R^F rangle$ is the by definition the subgroup $langle frf^{-1} : r in R, f in F rangle$ of $F$. It is not usually finitely generated. What exactly do you mean by "compute"?
$endgroup$
– Derek Holt
Apr 18 '17 at 9:48












$begingroup$
So is the normal closure of $R$ always equal to the kernel of $pi$? Because on one hand, $G$ is isomorphic to $F/ker(pi)$, while on the other hand $G$ is isomorphic to $F/N$ where $N$ is the normal closure of $R$.
$endgroup$
– W.Scott
Apr 18 '17 at 11:45






$begingroup$
So is the normal closure of $R$ always equal to the kernel of $pi$? Because on one hand, $G$ is isomorphic to $F/ker(pi)$, while on the other hand $G$ is isomorphic to $F/N$ where $N$ is the normal closure of $R$.
$endgroup$
– W.Scott
Apr 18 '17 at 11:45














$begingroup$
And for the $Z_7$ example, how can you show that $<R>=<R^F>$?
$endgroup$
– W.Scott
Apr 18 '17 at 11:49






$begingroup$
And for the $Z_7$ example, how can you show that $<R>=<R^F>$?
$endgroup$
– W.Scott
Apr 18 '17 at 11:49














$begingroup$
Yes, with your notation $N = {rm ker} pi$. In the $Z_7$ example $F$ is cyclic and hence abelian.
$endgroup$
– Derek Holt
Apr 18 '17 at 12:13




$begingroup$
Yes, with your notation $N = {rm ker} pi$. In the $Z_7$ example $F$ is cyclic and hence abelian.
$endgroup$
– Derek Holt
Apr 18 '17 at 12:13


















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Plaza Victoria

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Plaza Victoria Plaza Victoria en 2017 Otros nombres Plaza de La Victoria Localización El Almendral, Valparaíso, Chile Vías adyacentes Chacabuco, Molina, Plaza Victoria y Edwards Creación 1841 Metro Bellavista Ubicación 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 Coordenadas: 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 [editar datos en Wikidata] La Plaza Victoria , también llamada Plaza de la Victoria , [ 1 ] ​ es una plaza ubicada en el sector El Almendral del plan de la ciudad de Valparaíso, Chile, de importancia patrimonial, que incluye valiosas estatuas y especies vegetales. [ 2 ] ​ La plaza se creó como tal en la primera mitad del siglo XIX, tomando previamente los nombres de Plaza Nueva y Plaza de Orrego . Si bien desde sus inicios tuvo relevancia histórica, no fue sino a partir de fines de los años 1860 cuando comenzó a cobrar mayor importancia y se convirtió en uno de los principales centros sociales de l
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Puebla de Zaragoza

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Para otros usos de este término, véase Puebla (desambiguación). Heroica Puebla de Zaragoza Localidad De arriba a abajo de izquierda a derecha: Estrella de Puebla, Volcán Popocatépetl, Catedral de Puebla, Monumento a Ignacio Zaragoza, Estadio Cuauhtémoc, Angelópolis y Vista panorámica de la ciudad. Escudo Otros nombres: El relicario de América/ La ciudad de los Ángeles/ La Angelópolis Lema: Muy noble y muy leal ciudad de Puebla de los Ángeles Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en México Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en Puebla Mapa interactivo Coordenadas 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Coordenadas: 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Idioma oficial Español Entidad Localidad  • País   México  • Estado Puebla  • Municipio Puebla Presidente municipal Claudia Rivera Vivanc
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Musa

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Para el género de plantas, véase Musa (planta). Apolo y la Nueve Musas En la mitología griega, las musas (en griego antiguo μοῦσαι « mousai» ) son, según los escritores más antiguos, las divinidades inspiradoras de las Artes: cada una de ellas está relacionada con ramas artísticas y del conocimiento. Son hijas de Zeus y de Mnemósine, compañeras del séquito de Apolo, dios olímpico de la música y patrón de las bellas artes, quien tuvo romances con cada una de ellas, dejando descendientes. Bajaban a la tierra a susurrar ideas e inspirar a aquellos mortales que las invocaran. En la época más arcaica eran las ninfas inspiradoras de las fuentes, en las cuales eran adoradas. Finalmente, alrededor de los siglos VIII-VII a. C. [ 1 ] ​ prevaleció en todo el territorio de la Hélade la adoración de las nueve Musas, que son Calíope, Clío, Erato, Euterpe, Melpómene, Polimnia, Talía, Terpsícore y Urania. El culto a las musas era originalmente de Tracia y Beocia, y fueron de vital impo
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